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            	<h1 id="06-">06. 从尾到头打印链表</h1>
<p>难度：<strong>简单</strong></p>
<p>输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。</p>
<p>示例 1：</p>
<pre><code>输入：head = [1,3,2]
输出：[2,3,1]
</code></pre><p>限制：</p>
<p>0 &lt;= 链表长度 &lt;= 10000</p>
<p>来源：力扣（LeetCode）</p>
<p>链接：<a href="https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof">https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof</a></p>
<p>著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。</p>
<hr>
<p>解法1：递归，链表自带递归属性</p>
<pre><code class="lang-Java">class Solution {

    int[] res;

    int i = 0;

    int j = 0;

    public int[] reversePrint(ListNode head) {
        solve(head);
        return res;

    }

    //递归

    public void solve(ListNode head){
        //最后一次进入递归，根据i实例化数组，返回
        if(head == null){
            res = new int[i];
            return;
        }
        //确定数组长度
        i++;
        solve(head.next);
        //开始逐一赋值
        res[j++] = head.val;

    }
}
</code></pre>

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